/**
 * 给你一个由'1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 * 此外，你可以假设该网格的四条边均被水包围。
 *
 * 示例 1：
 * 输入：grid = [
 *   ["1","1","1","1","0"],
 *   ["1","1","0","1","0"],
 *   ["1","1","0","0","0"],
 *   ["0","0","0","0","0"]
 * ]
 * 输出：1
 *
 * 示例 2：
 * 输入：grid = [
 *   ["1","1","0","0","0"],
 *   ["1","1","0","0","0"],
 *   ["0","0","1","0","0"],
 *   ["0","0","0","1","1"]
 * ]
 * 输出：3
 *
 * 提示：
 * m == grid.length
 * n == grid[i].length
 * 1 <= m, n <= 300
 * grid[i][j] 的值为 '0' 或 '1'
 *
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/number-of-islands
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */
public class Q00200m {

    /**
     * 回溯
     * @param grid
     * @return
     */
    public int numIslands(char[][] grid) {
        int ans = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') { // 当前是1，开始向其周围寻找,将1染色为0
                    ans++;
                    dfs(grid, i, j);
                }
            }
        }
        return ans;
    }

    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') return;
        grid[i][j] = '0';
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}
